This should like a Poisson Distribution problem to me. What section are you in in your studies?

The reason I say Poisson is because you have an observed average over a unit of space: 2 trucks broken per 1 day on average. For Poisson we have where lambda is your average, and "x" is the number of events occurring that we wish to find the probability for . In this case, lamda is equal to 2 trucks/1 day = 2, and X refers to the number of trucks breaking down on a given day.

So for A: "a) What is the probability that on any given day the number of standby trucks is inadequate?" We have to, as you said, translate this into something that makes sense. On any given day, two trucks break down, and we have two in standby. What would need to happen for us to say that the number of trucks is inadequate? That would mean all standby trucks are gone, which would mean 2 trucks broke down, and then ANOTHER truck broke down. Therefore we need to find the probability that 3 or more trucks have broken down. However, our values of X can go on forever (well no, its limited by the number of trucks you actually have, but in theory. . .), so we can not calculate that. What we can calculate is the probability that less than 3 trucks broke down; so using Poisson: P(0)+P(1)+P(2) = ? However, that is the probability that less than three trucks break down. We want 3 or more, so we need to take the COMPLIMENT of that. Do so.

Use the same reasoning above with Poisson. In mathematics terms, and for the equation you have, what does it mean if no trucks break down? What would my X be. This question though is a little tricky. Originally we defined our lambda in terms of one day - 2 trucks/ 1 day. This is asking about TWO days. How would we adjust our lambda to get an average number of breaking trucks per two days?b) What is the probability that no delivery trucks will be inoperative during any two-day period?